Electrical – Gate drive for IRF540N using Arduino Uno

Method 1: Do nothing special...

Simple if the generated signal can be DC-biased by the signal generator, i.e. instead of -1.4 to 1.4 Volts, output a waveform of 1.0 to 3.8 Volts.

This signal can be directly used as digital input to an Arduino GPIO pin. For standard Arduino boards, V_cc is 5 Volts, while some clones and specific newer boards work at 3.3 Volts. For the 5 Volt case:

• GPIO port thresholds (from ATmega328 datasheet):
  • LOW is < 0.3 Vcc, i.e. < 1.5 Volts
  • HIGH is > 0.7 Vcc, i.e. > 3.5 Volts

Thus, raise the voltage floor of the square wave so it goes beyond these voltage levels at high and low, and it's all done.

Method 2: Use a comparator, or an Op Amp as comparator

This is as already suggested by Nick Alexeev in comments. Please note that the LF355N may not be suitable for this purpose: Minimum V_cc supported is +/- 5 Volts, i.e. 10 Volts in single supply configuration. You will need a (preferably) rail-to-rail output op-amp supporting single supply operation at V_cc of 5 Volts.

_{(from this web page, which has additional explanations)}

Clamp (or adjust at signal generator) the negative part of the incoming signal so it does not go below Ground potential. If the generator does not support DC biasing, a diode-based voltage clamp could be used, several suitable schematics show up on a web search.

Choose R1 and R2 such that the voltage divider provides a comparison threshold within the voltage high and low levels of the square wave, say 0.8 Volts. The output will be inverted, but will toggle between the supply and ground levels (or as close to the supply rails as the op amp chosen can drive its output) according to the input signal.

Method 3: Use an NPN transistor as a switch

A BJT designed for switching, such as the 2n2222, can be used for this purpose. This transistor is designed to withstand higher reverse bias voltages at the base than the -1.4 Volts that a 2.8 Volt peak to peak signal would have, so no additional care needs to be taken for the negative part of the cycle.

What is the better route?

• If the signal generator supports DC biasing, Method 1 is the obvious answer.
• If not, the simplest and least expensive solution would be Method 3.

The simulation looks basically correct. The gate of the MOSFET acts a small capacitor, so each channel's circuit looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

When the PWM output is high, it charges the gate through the diode. When it is low, the diode is reverse-biased and prevents the PWM output from discharging the gate. The gate does slowly discharge through the 1 MΩ resistor, but this discharge has only just started when the PWM goes high again.